Mời cao nhân vào giải 💁
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Gọi độ dài cạnh là a
Diện tích hình vuông là S=a^2
Độ dài bán kính hình tròn là a/2
Diện tích hình tròn là: (a/2)^2*3,14=a^2*3,14/4
Theo đề, ta có: a^2-a^2*3,14/4=0,215
=>a^2=1
=>a=1
=>Diện tích hình tròn là: 1^2*3,14/4=0,785cm2
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\(8x+50=2x+30\\ \Rightarrow8x+50-2x-30=0\\ \Rightarrow6x+20=0\\ \Rightarrow6x=-20\\ \Rightarrow x=-\dfrac{10}{3}\)
\(8 x + 50 = 2 x + 30\)
\(⇒8x+50−2x−30=0\)
\(⇒6x+20=0\)
\(⇒6x=−20\)
\(⇒x=-\frac{10}{3}\)
HT
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\(a,\left(-8,5\right)+16,35+\left(-4,5\right)-\left(-2,25\right)\\ =\left[\left(-8,5\right)+\left(-4,5\right)\right]+\left[16,35-\left(-2,25\right)\right]\\ =-13+18,6=5,6\\ b,5,63+\left(-2,75\right)-\left(-8,94\right)+9,06-15,25\\ =5,63-2,75+8,94+9,06-15,25\)
\(=5,63-\left(2,75+15,25\right)+\left(8,94+9,06\right)\\ =5,63-18+18\\ =5,63\)
a) (-8,5) + 16,35 + (-4,5) - (-2,25) = 5,6
b) 5,63 + (-2,75) - (-8,94) + 9,06 - 15,25 = 5,63
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\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{10}=\frac{y}{15}\\\frac{z}{4}=\frac{y}{5}\Leftrightarrow\frac{z}{12}=\frac{y}{15}\end{cases}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12};x+y-z=-39}\)
Tính chất dãy tỉ số bằng nhau:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=-\frac{39}{13}=-3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-3\Leftrightarrow x=-30\\\frac{y}{15}=-3\Leftrightarrow y=-45\\\frac{z}{12}=-3\Leftrightarrow z=-36\end{cases}}\)
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1 We should recycle plastic bags and jars
2 How can I get to the nearest bus stop?
3 We are going rafting so you should bring a towel
4 Because of feeling tired, I went to bed early
5 It is not difficult to recycle paper
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Bt1:
\(\dfrac{PTK_X}{PTK_{O_2}}=\dfrac{7}{8}\\ PTK_{O_2}=32\left(đvC\right)\)
\(\Rightarrow PTK_X=\dfrac{7}{8}.32=28\left(đvC\right)\)
\(\Rightarrow M_X=28\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow m_C=\%C.M_X=85,71\%.28=24\left(g\right)\\ m_H=m_X-m_C=28-24=4\left(g\right)\)
\(\Rightarrow n_C=\dfrac{m}{M}=\dfrac{24}{12}=2\left(mol\right)\\ n_H=\dfrac{m}{M}=\dfrac{4}{1}=4\left(\dfrac{g}{mol}\right)\)
\(CTHH:C_2H_4\)
Bt2:
\(\dfrac{PTK_A}{PTK_{O_2}}=5,625\\ PTK_{O_2}=32\left(đvC\right)\\ \Rightarrow PTK_A=5,625.32=180\left(đvC\right)\)
\(\Rightarrow M_A=180\left(đvC\right)\)
\(\Rightarrow m_C=\%C.M_A=40\%.180=72\left(g\right)\\ m_H=\%H.M_A=6,67\%.180=12\left(g\right)\\ m_O=m_A-m_C-m_H=180-72-12=96\left(g\right)\)
\(\Rightarrow n_C=\dfrac{m}{M}=\dfrac{72}{12}=6\left(mol\right)\\ n_H=\dfrac{m}{M}=\dfrac{12}{1}=12\left(mol\right)\\ n_O=\dfrac{m}{M}=\dfrac{96}{16}=6\left(mol\right)\)
\(CTHH:C_6H_{12}O_6\)
ulatr b này bt làm rùi mà thui CX củm mơn nhe 😆